Sum

Note that the Maclaurin series of $\cos = \displaystyle \sum_{n=0}^\infty \frac {(-1)^n x^{2n}}{(2n)!}$ . Therefore,

$\begin{aligned} S & = \sum_{n=0}^\infty \frac {(-1)^n \left(\left(\frac \pi 3\right)^{2n} +\left(\frac \pi 6\right)^{2n} \right)}{(2n)!} \\ & = \sum_{n=0}^\infty \frac {(-1)^n \left(\frac \pi 3\right)^{2n}}{(2n)!} + \sum_{n=0}^\infty \frac {(-1)^n \left(\frac \pi 6\right)^{2n}}{(2n)!} \\ & = \cos \frac \pi 3 + \cos \frac \pi 6 = \frac 12 + \frac {\sqrt 3} 2 = \frac {1+\sqrt 3}2 \end{aligned}$

$\implies a + b + c = 1+3+2 = \boxed{6}$

$e^{ia}$ = 1 + $ia$ - $\frac{a^2}{2!}$ - $\frac{ia^3}{3!}$ + $\frac{a^4}{4!}$ + ... , (equation 1)

Where $i^2$ = $-1$ ,

Now consider,

$e^{-ia}$ = 1 - $ia$ - $\frac{a^2}{2!}$ + $\frac{ia^3}{3!}$ + $\frac{a^4}{4!}$ - ... , (equation 2) ,

Now add (equation 1) and (equation 2) ,

$e^{ia}$ + $e^{-ia}$ = 2{1 - $\frac{a^2}{2!}$ + $\frac{a^4}{4!}$ -...} = 2 $cosa$ ,

Or directly, using maclaurin series , cosa = {1 - $\frac{a^2}{2!}$ + $\frac{a^4}{4!}$ -...} ,

$\large \sum_{n=0}^\infty \dfrac{ (-1)^n \left( \left( \frac \pi3\right)^{2n} + \left( \frac \pi6\right)^{2n} \right) }{(2n)!}$ = cos (60°) + cos(30°) ,

So answer is cos(60°) + cos(30°) = $\frac{1+✓3}{2}$

$a$ = 1 , $b$ = 3 , $c$ = 2,

$a+b+c$ = $6$ .